Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <cstdio>
#include <queue>
using namespace std;
 
vector<int> G[100001];
int S[100001], T[100001];
int Sc[100001], Tc[100001];
 
int main()
{
	int N, M, s, t;
	
	scanf("%d %d %d %d", &N, &M, &s, &t);
	
	for (int i = 0; i < M; i++){
		int src, dst;
		scanf("%d %d", &src, &dst);
		G[src].push_back(dst);
		G[dst].push_back(src);
	}
	
	for (int i = 1; i <= N; i++){
		S[i] = T[i] = 1000000000;
	}
	
	queue<int> q;
	
	S[s] = 0;
	
	for (q.push(s); q.size(); q.pop()){
		int x = q.front();
		Sc[S[x]]++;
		for (int i = 0; i < G[x].size(); i++){
			if (S[G[x][i]] > S[x] + 1){
				S[G[x][i]] = S[x] + 1; q.push(G[x][i]);
			}
		}
	}
	
	T[t] = 0;
	
	for (q.push(t); q.size(); q.pop()){
		int x = q.front();
		Tc[T[x]]++;
		for (int i = 0; i < G[x].size(); i++){
			if (T[G[x][i]] > T[x] + 1){
				T[G[x][i]] = T[x] + 1; q.push(G[x][i]);
			}
		}
	}
	
	long long ans = 0;
	for (int i = 0; S[t] - i - 2 >= 0; i++){
		ans += (long long)Sc[i] * Tc[S[t] - i - 2];
	}
	
	printf("%lld\n", ans);
	
	return (0);
}